Question: The base of a solid $S$ is the triangle enclosed by the line $x+y=1$, the $x$ -axis, and the $y$ -axis. $y$ $x$ $(0,1)$ $(1,0)$ ${x+y=1}$ Cross-sections perpendicular to the $x$ -axis are isosceles right triangles with the hypotenuse lying in the base. Determine the exact volume of solid $S$.
Let's graph the base of the solid. The thin orange rectangle depicts a representative cross-section sitting on the base. The length of the green segment is $y$. $y$ $x$ $y$ $(0,1)$ $(x,y)$ $(1,0)$ ${x+y=1}$ Since each cross-section is perpendicular to the $x$ -axis, the independent variable is $x$. If $A$ denotes the area of each cross-section as a function of $x$, the volume $V$ of solid $S$ is $ V=\int_a^b A(x) \,dx$. To determine the area $A$ as a function of $x$, first express $A$ in terms of $y$. Since the triangular cross-section rests on the rectangle pictured above, the length of the base of the triangle is $y$. Since the triangle is right isosceles with hypotenuse at the base, the height of the triangle is $y/2$. $\dfrac y2$ $y$ The area $A$ of the triangle is $A=\dfrac12\cdot y\cdot\dfrac y2=\dfrac14y^2$. What is $A$ as a function of $x$ ? The corner point $(x,y)$ of the rectangle lies on the line $x+y=1$. Let's rewrite the equation as $y=1-x$. Now we can express $A=\dfrac14y^2$ in terms of $x$ as $A(x)=\dfrac14(1-x)^2=\dfrac14(x-1)^2$. Can you express the volume $V$ of solid $S$ as a definite integral? Since $x$ goes from $0$ to $1$, the volume formula $ V=\int_a^b A(x) \,dx$ gives us the definite integral $\begin{aligned} V&=\int_0^1 \dfrac14(x-1)^2\,dx \\\\ &=\dfrac14\int_0^1 (x-1)^2\,dx \end{aligned}$ What is the value of the integral? $\begin{aligned} V&=\dfrac14\int_0^1 (x-1)^2\,dx \\\\ &=\dfrac14\left[\dfrac13(x-1)^3\right]_0^1 \\\\ &=\dfrac14\cdot\dfrac13\left[(1-1)^3-(0-1)^3\right] \\\\ &=\dfrac1{12}\left[(0)^3-(-1)^3\right] \\\\ &=\dfrac1{12} \end{aligned}$